03_Angular Momentum

1. Pure Rotation (Moving Axis)

When extending rotational kinematics from a fixed axis to a moving axis (e.g., a rolling object), we analyze the Center of Mass (CM).

• Kinematics of Pure Rolling:

  For an object of radius \(R\) rolling without slipping, the arc length \(s = R\theta\).

  • Velocity of CM: \(v_{cm} = \frac{ds}{dt} = R\frac{d\theta}{dt} = R\omega\)

  • Acceleration of CM: \(a_{cm} = \frac{dv_{cm}}{dt} = R\alpha\)

• Energy Conservation:

  When the point of contact remains static (no slipping), static friction does no work in pure rotation. Therefore, mechanical energy is conserved.

• Total Kinetic Energy (König's Theorem):

  Total kinetic energy \(K\) is the sum of translational kinetic energy of the CM and rotational kinetic energy relative to the CM:

  \[ K = K_{cm} + K' = \frac{1}{2} Mv_{cm}^2 + \frac{1}{2} I_{cm} \omega^2 \]

  For pure rotation (\(v_{cm} = \omega R\)), this simplifies to:

  \[ K = \frac{1}{2}\left(M + \frac{I_{cm}}{R^2}\right) v_{cm}^2 \]

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2. Torque and Angular Momentum

Recall the linear Impulse-Momentum Theorem: \(\vec{I} = \vec{P}_f - \vec{P}_i\). We can define a rotational analogue.

Definitions

• Torque: \(\vec{\tau} = \vec{r} \times \vec{F}\)

• Angular Momentum: \(\vec{L} = \vec{r} \times \vec{p}\)

Derivation of Rotational Dynamics

Taking the time derivative of angular momentum:

\[ \vec{\tau} = \vec{r} \times \frac{d\vec{p}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) - \frac{d\vec{r}}{dt} \times \vec{p} \]

Since \(\frac{d\vec{r}}{dt} \times \vec{p} = \vec{v} \times (m\vec{v}) = 0\), we get the rotational analogue to Newton's Second Law:

\[ \vec{\tau} dt = d\vec{L} \]

Key Notes:

1. Both \(\vec{\tau}\) and \(\vec{L}\) depend on the choice of origin.

2. The origin for calculating \(\vec{\tau}\) and \(\vec{L}\) must be the same.

3. Any chosen origin is mathematically valid.

System of Particles

For a system of particles, total angular momentum is \(\vec{L} = \sum_i \vec{L}_i\) and total external torque is \(\vec{\tau}_{ext} = \sum_i \vec{\tau}_{i, ext}\).

\[ \vec{\tau}_{ext} dt = d\vec{L} \]

Note: Internal torques cancel out. For example, the internal torque between two particles is \(\vec{\tau}_{int} = \vec{r}_1 \times \vec{F}_{12} + \vec{r}_2 \times \vec{F}_{21} = (\vec{r}_1 - \vec{r}_2) \times \vec{F}_{12} = 0\) (since the distance vector is parallel to the force vector). Only external forces need to be considered.

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3. The Moment of Inertia Tensor

For a rigid body rotating around a fixed axis:

\[ \vec{L} = \sum_i \vec{r}_i \times m_i \vec{v}_i = \sum_i \vec{r}_i \times m_i (\vec{\omega} \times \vec{r}_i) \]

Applying the vector triple product expansion:

\[ \vec{L} = \sum_i m_i \left[ r_i^2 \vec{\omega} - (\vec{r}_i \cdot \vec{\omega})\vec{r}_i \right] \]

By expanding the vectors into Cartesian components \(\vec{\omega} = (w_x, w_y, w_z)^T\) and \(\vec{r}_i = (x_i, y_i, z_i)^T\), we can express this as a matrix multiplication:

\[ \vec{L} = \sum_i m_i \begin{pmatrix} (y_i^2+z_i^2) & -x_i y_i & -x_i z_i \\ -y_i x_i & (z_i^2+x_i^2) & -y_i z_i \\ -z_i x_i & -z_i y_i & (x_i^2+y_i^2) \end{pmatrix} \begin{pmatrix} w_x \\ w_y \\ w_z \end{pmatrix} \]

This establishes the relationship \(\vec{L} = I \vec{\omega}\), where \(I\) is the Moment of Inertia Tensor:

\[ I = \sum_i m_i (r_i^2 \mathbf{1} - \vec{r}_i \otimes \vec{r}_i) \]

(Note: \(\vec{A} \otimes \vec{B}\) denotes the dyadic/direct product where \((\vec{A} \otimes \vec{B})_{mn} \equiv A_m B_n\).)

Example: Uniform Ball

For a uniform sphere, the diagonal elements are all \(I_0 = \frac{2}{5} MR^2\).

The off-diagonal elements (e.g., \(I_{xy} = \int dx dy dz (-xy) \rho\)) integrate to zero due to odd symmetry in \(x\) and \(y\).

Thus, if the z-axis is fixed, \(L_z = I_0 \omega_z\).

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4. Conservation of Angular Momentum

When net external torque is zero (\(\vec{\tau}_{ext} = 0\)), then \(\frac{d\vec{L}}{dt} = 0\), meaning \(\vec{L}\) is constant.

Example: Kepler's Second Law

A planet orbits the Sun under a central gravitational force \(\vec{F}_g\).

\[ \vec{\tau} = \vec{r} \times \vec{F}_g = 0 \implies \vec{L} = \vec{r} \times \vec{p} = \text{constant} \]

The area swept by \(\vec{r}\) in unit time is \(dA = \frac{1}{2} |\vec{r} \times \vec{v} dt|\).

\[ \frac{dA}{dt} = \frac{1}{2} |\vec{r} \times \vec{v}| = \frac{|\vec{L}|}{2m} = \text{constant} \]

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5. Dynamics in the Center of Mass (CM) Frame

• \(\vec{r} = \vec{r}_{cm} + \vec{r}'\):

• Torque Separation:

  \[ \vec{\tau}_{ext} = \vec{r}_{cm} \times \left(\sum_i \vec{F}_{i,ext}\right) + \sum_i \vec{r}_i' \times \vec{F}_{i,ext} \]

  This splits torque into the torque from all external forces acting on the CM, plus the torques relative to the CM. (Note: Torque from gravity relative to CM is 0).

• Angular Momentum Separation:

  \[ \vec{L} = M\vec{r}_{cm} \times \vec{v}_{cm} + \sum_i m_i \vec{r}_i' \times \vec{v}_i' \]

  This splits angular momentum into the orbital angular momentum of the CM, plus the spin angular momentum relative to the CM.

Combined Formalism:

\[ \sum_i \vec{r}_i' \times \vec{F}_{i,ext} = \frac{d}{dt} \left(\sum_i m_i \vec{r}_i' \times \vec{v}_i'\right) \]

\[ \vec{r}_{cm} \times \left(\sum_i \vec{F}_{i,ext}\right) = \frac{d}{dt} (M\vec{r}_{cm} \times \vec{v}_{cm}) \]

For any closed, isolated system, the following are strictly conserved:

1. Energy: \(K_i + U_i = K_f + U_f\)

2. Linear Momentum: \(\vec{P}_i = \vec{P}_f\)

3. Angular Momentum: \(\vec{L}_i = \vec{L}_f\)

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6. Comprehensive Examples

A. The Rolling Ball on an Incline

Find the CM acceleration (\(a_{cm}\)) of a ball rolling down an incline of angle \(\theta\).

• Method 1: Momentum & Relative Angular Momentum

  • Translation: \(Mg \sin\theta - f_s = M a_{cm}\)

  • Rotation: \(f_s R = I_{cm} \alpha\)

  • Constraint (pure rotation): \(a_{cm} = \alpha R\)

  • Solving yields: \(a_{cm} = \frac{g \sin\theta}{1 + I_{cm}/(MR^2)}\)

• Method 2: Parallel Axis Theorem

  Take the torque about the bottom contact point (where normal force \(\vec{n}\) and friction \(\vec{f}_s\) exert no torque).

  • \(\tau_{ext} = Mg R \sin\theta\)

  • \(I = I_{cm} + MR^2\)

  • \(Mg R \sin\theta = (I_{cm} + MR^2) \frac{a_{cm}}{R} \implies a_{cm} = \frac{g \sin\theta}{1 + I_{cm}/(MR^2)}\)

B. Elastic Collision (Ball and Stick)

A ball (mass \(m_b\), initial velocity \(v_{bi}\)) strikes a freely pivoting stick (mass \(M\)) at distance \(r\) from the pivot. Find final velocities \(v_{bf}\) (ball), \(v_s\) (stick CM), and \(\omega\) (stick angular velocity).

• schematic diagram from the hand-written notes of Mr. ztWang

Set up a system of three conservation equations:

1. Momentum: \(m_b v_{bi} = m_b v_{bf} + M v_s\)

2. Angular Momentum: \(m_b v_{bi} r = m_b v_{bf} r + I \omega\)

3. Kinetic Energy: \(\frac{1}{2} m_b v_{bi}^2 = \frac{1}{2} m_b v_{bf}^2 + \frac{1}{2} M v_s^2 + \frac{1}{2} I \omega^2\)

C. Rotation with Slipping

A solid cylinder (\(I = \frac{1}{2}MR^2\)) is placed on a surface with initial forward velocity \(V_0\) and initial backspin \(\omega_0\). Kinetic friction \(\mu Mg\) acts to normalize the motion. Find the time \(t\) when slipping stops, and the final pure-rolling velocity \(V_{cm}\).

1. Linear deceleration: \(V_{cm} = V_0 - \mu gt\)

2. Angular acceleration: \(\omega = -\omega_0 + \frac{\mu MgR}{I} t\)

3. Pure rolling condition: \(V_{cm} = \omega R\)

Solving the system:

\[ t = \frac{V_0 + \omega_0 R}{\mu g (1 + \frac{MR^2}{I})} = \frac{V_0 + \omega_0 R}{3 \mu g} \]

\[ V_{cm} = \frac{1}{3} (2V_0 - \omega_0 R) \]

D. Precession of a Gyroscope

A fast-spinning gyroscope tilted at angle \(\theta\) experiences a gravitational torque.

• Angular momentum is dominated by the fast spin: \(\vec{L} \approx \sum_i \vec{r}_i' \times \vec{p}_i'\) (CM orbital term \(\vec{r}_{cm} \times \vec{p}_{cm} \approx 0\)).

• Torque: \(\vec{\tau} = \vec{r}_{cm} \times M\vec{g} \implies \tau = Mg r \sin\theta = \frac{dL}{dt}\)

• From geometry, a small change in angle \(d\phi\) means \(dL = L \sin\theta d\phi\).

• Substituting: \(Mg r \sin\theta = L \sin\theta \frac{d\phi}{dt}\)

• Precession angular velocity: \(\frac{d\phi}{dt} = \frac{Mgr}{L}\)